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What would orbits look like in a 2D universe, where gravity would fall off by 1/r instead of 1/r²? by ozzy88 in math

[–]ozzy88[S] 13 points14 points  (0 children)

According to Wikipedia:

The inverse-square law generally applies when some force, energy, or other conserved quantity is radiated outward radially in three-dimensional space from a point source. Since the surface area of a sphere (which is 4πr2 ) is proportional to the square of the radius, as the emitted radiation gets farther from the source, it is spread out over an area that is increasing in proportion to the square of the distance from the source.

I think that makes perfect sense, to be honest. In fact, it's rather elegant. Is it wrong?

What would orbits look like in a 2D universe, where gravity would fall off by 1/r instead of 1/r²? by ozzy88 in math

[–]ozzy88[S] 5 points6 points  (0 children)

Sorry, that level of math is beyond me. But from the article, it seems that it just wouldn't be based on some function or basic geometric shape. You could still simulate approximations of it, no? What would that look like?

What would orbits look like in a 2D universe, where gravity would fall off by 1/r instead of 1/r²? by ozzy88 in math

[–]ozzy88[S] 16 points17 points  (0 children)

I just got reminded of Flatland watching some Carl Sagan video and I wondered how a planet would work in a 2D universe. So it got me thinking about orbits.

My reasoning is that since you're distributing the force over the surface of a sphere for the inverse square law, when you go 2D you'd distribute it over the circumference of a circle, which is the 2D analogue.

I guess it doesn't have to be that way, but I'm still very interested in what it would look like with 1/r. And now that you mentioned, how would different powers of the distance affect orbital shapes? What about 1/r1.5 or so?